3.27. Let R be a ring, possibly non commutative, in which xy = 0 implies x = 0 or y = 0. If a, b R and an = bn and am = fm for two relatively prime positive integers m and n, prove that a = b. Solution: (m, n) = 1, implies that, there exists u, k = Z such that mu+nk = 1. Since m, n ≥ 0 either u ≥ 0 or k ≥ 0. Assume that u > 0. Then munk = 1 where k > 0. Hence mu = nk + 1. Now, let
3.27. Let R be a ring, possibly non commutative, in which xy = 0 implies x = 0 or y = 0. If a, b R and an = bn and am = fm for two relatively prime positive integers m and n, prove that a = b. Solution: (m, n) = 1, implies that, there exists u, k = Z such that mu+nk = 1. Since m, n ≥ 0 either u ≥ 0 or k ≥ 0. Assume that u > 0. Then munk = 1 where k > 0. Hence mu = nk + 1. Now, let
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter6: More On Rings
Section6.3: The Characteristic Of A Ring
Problem 7E
Related questions
Question
![3.27. Let R be a ring, possibly non commutative, in which xy 0
implies x = 0 or y = 0. If a,b ≤ R and an = bn and am
a" = fm for two
relatively prime positive integers m and n, prove that a = = b.
Solution: (m, n) = 1, implies that, there exists u, k € Z such that
mu+nk = 1. Since m, n ≥ 0 either u ≥ 0 or k ≥ 0. Assume that u > 0.
Then mu - nk = 1 where k > 0. Hence mu = nk + 1. Now, let
mu
a = (am)u = (fm)u = fmu
ank+1
=amu = a
fnk+1
=
aank = a(an)k = a(bn)k = abnk
=
Hence
bb nk abrk implies that (b − a)bnk = 0. Then either bnk = 0 or
b = a. The first possibility fnk = 0 is impossible if b ‡ 0. One can see
this easily that, if t is the smallest positive integer such that bt-1 0,
but bt = 0, then 0 = b = bt-¹.b 0. This implies that the assumption
bt = 0 is impossible, whenever b 0. Hence b
=
Request explain the
highlighted portion
If u or k >=0, and
we assume u>0,
how do we say then
k>0??](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F8317a119-7a4a-45f0-946d-0fee47bc0454%2Fuyq6gsm_processed.png&w=3840&q=75)
Transcribed Image Text:3.27. Let R be a ring, possibly non commutative, in which xy 0
implies x = 0 or y = 0. If a,b ≤ R and an = bn and am
a" = fm for two
relatively prime positive integers m and n, prove that a = = b.
Solution: (m, n) = 1, implies that, there exists u, k € Z such that
mu+nk = 1. Since m, n ≥ 0 either u ≥ 0 or k ≥ 0. Assume that u > 0.
Then mu - nk = 1 where k > 0. Hence mu = nk + 1. Now, let
mu
a = (am)u = (fm)u = fmu
ank+1
=amu = a
fnk+1
=
aank = a(an)k = a(bn)k = abnk
=
Hence
bb nk abrk implies that (b − a)bnk = 0. Then either bnk = 0 or
b = a. The first possibility fnk = 0 is impossible if b ‡ 0. One can see
this easily that, if t is the smallest positive integer such that bt-1 0,
but bt = 0, then 0 = b = bt-¹.b 0. This implies that the assumption
bt = 0 is impossible, whenever b 0. Hence b
=
Request explain the
highlighted portion
If u or k >=0, and
we assume u>0,
how do we say then
k>0??
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